I love how perfectly clear you are, both with your words and your symbols. So easy to follow along, and your writing is very tidy! Would've loved to have you as a math teacher in uni! Any courses you teach?
Wouldn't it be simpler to use partial fractions on the integral (x-1)/(x^2-x-1) at 1:58? We know the roots of the denominator are the golden ratio and its conjugate. You get a very simple sum that already includes the golden ratio.The ln of the golden ratio comes out naturally (no pun intended). I mean, I can't think of anything more enjoyable than to integrate sec(t) first thing in the morning and figure out the value of inverse trigonometric functions, but still....
That's how I did it before watching the vid. And like he said, using partial fractions or trig substitutions is a coin flip assuming you are fluent in trig substitutions. My trig is really rusty, so I prefer partial fractions which is just a bunch of algebra. Plus, trig substitutions get really messy IMO when dealing with definite integrals and having to worry about domains.
I did this too. But on the other hand, one of the beautiful things about doing non-trivial integrals (and the same applies to derivatives as well) is that no matter how you do it, you get the same answer. And sometimes (with derivatives and indefinite integrals), if you try it multiple ways, you get expressions that don't look alike but actually turn out to be equivalent -- or in the case of indefinite integrals, differ by a constant. And in so doing, you can learn something very interesting.
This video uses more secans than I have used in my entire Gymnasium-level and University-level math education in Germany. I am bamboozled that people still use that...
I tried to solve the integral 1/(5/4-u^2)du using partial fractions and got a different result. Integral 1/(5/4-u^2)du = (1/sqrt(5))*Integral [1/(sqrt(5)/2+u) + 1/(sqrt(5)/2-u)]du = (1/sqrt(5))*[ ln(sqrt(5)/2+u) - ln(sqrt(5)/2-u)] + c For the limits from 0 to 1/2 I get = (1/sqrt(5))*[ ln((sqrt(5)+1)/2) - ln(sqrt(5)/2) - ln((sqrt(5)-1)/2) + ln(sqrt(5)/2)] = (1/sqrt(5))*ln[(sqrt(5)+1)/(sqrt(5)-1)] = (1/sqrt(5))*ln[(3+sqrt(5))/2] and this is not the result (2/sqrt(5))*ln[(sqrt(5)+1)/2)] from the video. Does anyone find my mistake?
Doesn't this make finding the bounds of integration more annoying? TBH I think partial fractions are a lot easier with definite integrals for this reason...
I would say the very first step is to show that the integral is actually defined between 0 and 1 which boils down to showing the denominator has no 0 between 0 and 1 which is true and is easy to show but that should be the very first step you should do for a calc exam to make sure you're not doing any nonsense. Also if someone could tell me how the roots of x^3-x^2-1 and x^2-x-1 are actually related i would gladly learn something new
He did show that the denominator has no 0 between 0 and 1. In fact, it's stated right on the board at the start of the video. The positive root of the denominator is the golden ratio, which is greater than 1 (and as this being the 'positive root' implies, the other root is negative). So neither root are between 0 and 1. As far as the super golden ratio, I'm not familiar with the topic myself, but he has a video from about a month ago about it.
No, he pronounces Phi ("fee") correctly but did not go for the right pronunciation of Psi ("psee", with a non-silent p). Due to the English vowels being royally fucked up and all over the place, it is one of the few languages that is butchering the Greek sounds beyong recognition. Everybody else in Europe, even the French and the Russians, try to imitate the Greek pronunciation when using Greek letters.
I love how perfectly clear you are, both with your words and your symbols. So easy to follow along, and your writing is very tidy! Would've loved to have you as a math teacher in uni! Any courses you teach?
Anything related to golden ratio is pure gold :) Great job, Michael!
pure god/good
No the golden ratio is overrated
@@lucid_ It's abused every single video it's super boring now
Wouldn't it be simpler to use partial fractions on the integral (x-1)/(x^2-x-1) at 1:58? We know the roots of the denominator are the golden ratio and its conjugate. You get a very simple sum that already includes the golden ratio.The ln of the golden ratio comes out naturally (no pun intended). I mean, I can't think of anything more enjoyable than to integrate sec(t) first thing in the morning and figure out the value of inverse trigonometric functions, but still....
That's how I did it before watching the vid. And like he said, using partial fractions or trig substitutions is a coin flip assuming you are fluent in trig substitutions. My trig is really rusty, so I prefer partial fractions which is just a bunch of algebra. Plus, trig substitutions get really messy IMO when dealing with definite integrals and having to worry about domains.
I did this too. But on the other hand, one of the beautiful things about doing non-trivial integrals (and the same applies to derivatives as well) is that no matter how you do it, you get the same answer. And sometimes (with derivatives and indefinite integrals), if you try it multiple ways, you get expressions that don't look alike but actually turn out to be equivalent -- or in the case of indefinite integrals, differ by a constant. And in so doing, you can learn something very interesting.
Ngl I was actually with him until he got to the trig substitution. I know some of the trig identities but not all of them. Threw me off a bit
Excellent video, Michael, and superbly narrated. Love your programming.
Dr penn never skipped integral day hhh😂
Cool video. I'll be going back and be watching the videos that I missed.
It's interesting how PV numbers raised to high powers are almost integers.
@5:41 I would have just assumed the anti derivative was of the form b* artanh (x/a) where a^2 = 5/4 and b is some constant
sqrt(5)/2*tanh(t) would work as well.
Having my breakfast while warming up My Neurons. Math is an excelent anti-aging Neuronal system pills.
-Thanks
This video uses more secans than I have used in my entire Gymnasium-level and University-level math education in Germany.
I am bamboozled that people still use that...
1/2+1/sqrt5ln(sqrt5+1/sqrt5-1)
46 minit slate
Who else got progressively nervous that Michael forgot the 2/sqr(5)?
Well I do have a simpler method but how do I share !? If I add a link RUclips simply will put down my comment!
Without graphing it, I'm not seeing how we know that is an odd function. It's because we have an odd function divided by an even function?
yep
an odd function divided by an even function is always odd
Golden ratio is the most irrational number. Can someone please enlighten me about the degree of irrationality and its applications?
Cooool!
You: Phi
me: 😊
You: Psy
me: 😒
I tried to solve the integral 1/(5/4-u^2)du using partial fractions and got a different result.
Integral 1/(5/4-u^2)du = (1/sqrt(5))*Integral [1/(sqrt(5)/2+u) + 1/(sqrt(5)/2-u)]du
= (1/sqrt(5))*[ ln(sqrt(5)/2+u) - ln(sqrt(5)/2-u)] + c
For the limits from 0 to 1/2 I get
= (1/sqrt(5))*[ ln((sqrt(5)+1)/2) - ln(sqrt(5)/2) - ln((sqrt(5)-1)/2) + ln(sqrt(5)/2)]
= (1/sqrt(5))*ln[(sqrt(5)+1)/(sqrt(5)-1)]
= (1/sqrt(5))*ln[(3+sqrt(5))/2]
and this is not the result (2/sqrt(5))*ln[(sqrt(5)+1)/2)] from the video.
Does anyone find my mistake?
There is no problem this is the same answer
3+sqrt5/2 = ((1+sqrt5)/2)^2 and by the log rule you get the same answer
@@yoav613
Many Thanks!
I see, using the logarithm laws I get the same result
I like your videos more when I know how to solve it by myself
9:52 ratio
That was fun.
(Also, I did it by myself, so I feel proudish)
Thank you, professor.
I don’t know why you don’t see the hyperbolic tangent and insist on the trigonometric change of variables
Hi, I don't know why but hyperbolic function are always underrated :(. I use them a lot for sostitution
Doesn't this make finding the bounds of integration more annoying?
TBH I think partial fractions are a lot easier with definite integrals for this reason...
@@TheEternalVortex42 The partial fractions will end up with a log of fraction in x which is exactly the definition of hyperbolic tangent :)
@@TheEternalVortex42 yes, but you can also solve first the indefinite integral then go back to x and substitute
Nice!
Cool video! I just wished you spoke English more often in the video than math for better understanding.
I would say the very first step is to show that the integral is actually defined between 0 and 1 which boils down to showing the denominator has no 0 between 0 and 1 which is true and is easy to show but that should be the very first step you should do for a calc exam to make sure you're not doing any nonsense. Also if someone could tell me how the roots of x^3-x^2-1 and x^2-x-1 are actually related i would gladly learn something new
He did show that the denominator has no 0 between 0 and 1. In fact, it's stated right on the board at the start of the video. The positive root of the denominator is the golden ratio, which is greater than 1 (and as this being the 'positive root' implies, the other root is negative). So neither root are between 0 and 1.
As far as the super golden ratio, I'm not familiar with the topic myself, but he has a video from about a month ago about it.
You could have used king property and the numerator will vanish by adding the two integrals
ikr
The lack of rigor is really a bad example.
Hi Dr.!
This should be called Giorno’s Integral
I love you michael❤
ничего не понял
Cool 🍺🍺🍻.
вы слишком разжёвываете
I got the result = -4.78 appox.
Oh, come on. Phi is pronounced "fee" and psi rhymes with "pie"?! How silly is that?
No, he pronounces Phi ("fee") correctly but did not go for the right pronunciation of Psi ("psee", with a non-silent p).
Due to the English vowels being royally fucked up and all over the place, it is one of the few languages that is butchering the Greek sounds beyong recognition. Everybody else in Europe, even the French and the Russians, try to imitate the Greek pronunciation when using Greek letters.
2nd thank you